3390: [Usaco2004 Dec]Bad Cowtractors牛的报复
Time Limit: 1 Sec Memory Limit: 128 MBDescription
Input
Output
Sample Input
1 2 3
1 3 7
2 3 10
2 4 4
2 5 8
3 4 6
3 5 2
4 5 17
Sample Output
连接4和5,2和5,2和3,1和3,花费17+8+10+7=42
HINT
#include<map> #include<cmath> #include<queue> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; #define ll long long #define N 1010 #define M 20010 inline int rd() { int x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } struct qaz{int a,b;ll v;}e[M]; bool cmp(qaz a,qaz b){return a.v>b.v;} int cnt,fa[N]; int findf(int x){return x==fa[x]?x:fa[x]=findf(fa[x]);} int n,m,x,y; ll ans; int main() { n=rd();m=rd(); int i,j,f1,f2; for(i=1;i<=n;i++) fa[i]=i; for(i=1;i<=m;i++){e[i].a=rd();e[i].b=rd();e[i].v=rd();} sort(e+1,e+m+1,cmp); for(int i=1;i<=m;i++) { x=findf(e[i].a);y=findf(e[i].b); if(x!=y) { ans+=e[i].v; fa[y]=x; } } bool f=0;x=findf(1); for(int i=1;i<=n;i++) if(findf(i)!=x){f=1;break;} f?puts("-1"):printf("%lld\n",ans); return 0; }
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